Chapter 23 Electric Potential

Chapter 23 Electric Potential
Chapter 23 opener. We are used to voltage in our lives—a 12-volt car battery, 110 V or 220 V at home, 1.5 volt flashlight batteries, and so on. Here we see a Van de Graaff generator, whose voltage may reach 50,000 V or more. Voltage is the same as electric potential difference between two points. Electric potential is defined as the potential energy per unit charge. The children here, whose hair stands on end because each hair has received the same sign of charge, are not harmed by the voltage because the Van de Graaff cannot provide much current before the voltage drops. (It is current through the body that is harmful, as we will see later.)

23-1 Electrostatic Potential Energy and Potential Difference
The electrostatic force is conservative – potential energy can be defined. Change in electric potential energy is negative of work done by electric force: Figure Work is done by the electric field in moving the positive charge q from position a to position b.

Electric potential is defined as potential energy per unit charge: Unit of electric potential: the volt (V): 1 V = 1 J/C.

Only changes in potential can be measured, allowing free assignment of V = 0:

Conceptual Example 23-1: A negative charge. Suppose a negative charge, such as an electron, is placed near the negative plate at point b, as shown here. If the electron is free to move, will its electric potential energy increase or decrease? How will the electric potential change? Figure Central part of Fig. 23–1, showing a negative point charge near the negative plate, where its potential energy (PE) is high. Example 23–1. Solution: The electron will move towards the positive plate if released, thereby increasing its kinetic energy. Its potential energy must therefore decrease. However, it is moving to a region of higher potential V; the potential is determined only by the existing charge distribution and not by the point charge. U and V have different signs due to the negative charge.

Analogy between gravitational and electrical potential energy: Figure (a) Two rocks are at the same height. The larger rock has more potential energy. (b) Two charges have the same electric potential. The 2Q charge has more potential energy.

Electrical sources such as batteries and generators supply a constant potential difference. Here are some typical potential differences, both natural and manufactured:

Example 23-2: Electron in CRT. Suppose an electron in a cathode ray tube is accelerated from rest through a potential difference Vb – Va = Vba = V. (a) What is the change in electric potential energy of the electron? (b) What is the speed of the electron (m = 9.1 × kg) as a result of this acceleration? Solution: a. The change in potential energy is qV = -8.0 x J. b. The change in potential energy is equal to the change in kinetic energy; solving for the final speed gives v = 4.2 x 107 m/s.

23-2 Relation between Electric Potential and Electric Field
The general relationship between a conservative force and potential energy: Substituting the potential difference and the electric field: Figure To find Vba in a nonuniform electric field E, we integrate E·dl from point a to point b.

The simplest case is a uniform field:

Example 23-3: Electric field obtained from voltage. Two parallel plates are charged to produce a potential difference of 50 V. If the separation between the plates is m, calculate the magnitude of the electric field in the space between the plates. Solution: E = V/d = 1000 V/m.

Example 23-4: Charged conducting sphere. Determine the potential at a distance r from the center of a uniformly charged conducting sphere of radius r0 for (a) r > r0, (b) r = r0, (c) r < r0. The total charge on the sphere is Q. Solution: The electric field outside a conducting sphere is Q/(4πε0r2). Integrating to find the potential, and choosing V = 0 at r = ∞: a. V = Q/4πε0r. b. V = Q/4πε0r0. c. V = Q/4πε0r0 (the potential is constant, as there is no field inside the sphere).

The previous example gives the electric potential as a function of distance from the surface of a charged conducting sphere, which is plotted here, and compared with the electric field: Figure (a) E versus r, and (b) V versus r, for a uniformly charged solid conducting sphere of radius r0 (the charge distributes itself on the surface); r is the distance from the center of the sphere.

Example 23-5: Breakdown voltage. In many kinds of equipment, very high voltages are used. A problem with high voltage is that the air can become ionized due to the high electric fields: free electrons in the air (produced by cosmic rays, for example) can be accelerated by such high fields to speeds sufficient to ionize O2 and N2 molecules by collision, knocking out one or more of their electrons. The air then becomes conducting and the high voltage cannot be maintained as charge flows. The breakdown of air occurs for electric fields of about 3.0 × 106 V/m. (a) Show that the breakdown voltage for a spherical conductor in air is proportional to the radius of the sphere, and (b) estimate the breakdown voltage in air for a sphere of diameter 1.0 cm. Solution: a. Combining the equations for the field and the potential gives V = r0E. b. Substituting gives V = 15,000 V.

23-3 Electric Potential Due to Point Charges
To find the electric potential due to a point charge, we integrate the field along a field line: Figure We integrate Eq. 23–4a along the straight line (shown in black) from point a to point b. The line ab is parallel to a field line.

Setting the potential to zero at r = ∞ gives the general form of the potential due to a point charge: Figure Potential V as a function of distance r from a single point charge Q when the charge is positive. Figure Potential V as a function of distance r from a single point charge Q when the charge is negative.

Example 23-6: Work required to bring two positive charges close together. What minimum work must be done by an external force to bring a charge q = 3.00 μC from a great distance away (take r = ∞) to a point m from a charge Q = 20.0 µC? Solution: The work is equal to the change in potential energy; W = 1.08 J. Note that the field, and therefore the force, is not constant.

Example 23-7: Potential above two charges. Calculate the electric potential (a) at point A in the figure due to the two charges shown, and (b) at point B. Solution: The total potential is the sum of the potential due to each charge; potential is a scalar, so there is no direction involved, but we do have to keep track of the signs. a. V = 7.5 x 105 V b. V = 0 (true at any point along the perpendicular bisector)

23-4 Potential Due to Any Charge Distribution
The potential due to an arbitrary charge distribution can be expressed as a sum or integral (if the distribution is continuous): or

Example 23-8: Potential due to a ring of charge. A thin circular ring of radius R has a uniformly distributed charge Q. Determine the electric potential at a point P on the axis of the ring a distance x from its center. Solution: Each point on the ring is the same distance from point P, so the potential is just that of a charge Q a distance (R2 + x2)1/2 from point P.

Example 23-9: Potential due to a charged disk. A thin flat disk, of radius R0, has a uniformly distributed charge Q. Determine the potential at a point P on the axis of the disk, a distance x from its center. Solution: Consider the disk to be made up of infinitely thin rings, each at a radius R with a thickness dR. Each ring then carries a charge dq = 2QR dR/R02. Integrating to find V then gives the solution in the text.

23-5 Equipotential Surfaces
An equipotential is a line or surface over which the potential is constant. Electric field lines are perpendicular to equipotentials. The surface of a conductor is an equipotential. Figure Equipotential lines (the green dashed lines) between two oppositely charged parallel plates. Note that they are perpendicular to the electric field lines (solid red lines).

Example 23-10: Point charge equipotential surfaces. For a single point charge with Q = 4.0 × 10-9 C, sketch the equipotential surfaces (or lines in a plane containing the charge) corresponding to V1 = 10 V, V2 = 20 V, and V3 = 30 V. Solution: Equipotential surfaces are spheres surrounding the charge; radii are shown in the figure (in meters).

Equipotential surfaces are always perpendicular to field lines; they are always closed surfaces (unlike field lines, which begin and end on charges). Figure Equipotential lines (green, dashed) are always perpendicular to the electric field lines (solid red) shown here for two equal but oppositely charged particles.

A gravitational analogy to equipotential surfaces is the topographical map – the lines connect points of equal gravitational potential (altitude). Figure A topographic map (here, a portion of the Sierra Nevada in California) shows continuous contour lines, each of which is at a fixed height above sea level. Here they are at 80 ft (25 m) intervals. If you walk along one contour line, you neither climb nor descend. If you cross lines, and especially if you climb perpendicular to the lines, you will be changing your gravitational potential (rapidly, if the lines are close together).

23-6 Electric Dipole Potential
The potential due to an electric dipole is just the sum of the potentials due to each charge, and can be calculated exactly. For distances large compared to the charge separation: Figure Electric dipole. Calculation of potential V at point P.

23-7 E Determined from V If we know the field, we can determine the potential by integrating. Inverting this process, if we know the potential, we can find the field by differentiating: This is a vector differential equation; here it is in component form:

23-7 E Determined from V Example 23-11: E for ring and disk.
Use electric potential to determine the electric field at point P on the axis of (a) a circular ring of charge and (b) a uniformly charged disk. Solution: a. Just do the derivatives of the result of Example 23-8; the only nonzero component is in the x direction. b. Same as (a); use the result of Example 23-9.

23-8 Electrostatic Potential Energy; the Electron Volt
The potential energy of a charge in an electric potential is U = qV. To find the electric potential energy of two charges, imagine bringing each in from infinitely far away. The first one takes no work, as there is no field. To bring in the second one, we must do work due to the field of the first one; this means the potential energy of the pair is:

One electron volt (eV) is the energy gained by an electron moving through a potential difference of one volt: 1 eV = 1.6 × J. The electron volt is often a much more convenient unit than the joule for measuring the energy of individual particles.

Example 23-12: Disassembling a hydrogen atom. Calculate the work needed to “disassemble” a hydrogen atom. Assume that the proton and electron are initially separated by a distance equal to the “average” radius of the hydrogen atom in its ground state, × m, and that they end up an infinite distance apart from each other. Solution: After separation, the total energy – potential plus kinetic – is zero. The energy (work) needed to separate the proton and electron is the negative of the potential energy plus the kinetic energy of the atom before disassembly. The potential energy is eV (convert from joules), and the kinetic energy of the electron in its orbit is 13.6 eV (calculate its speed from F = ma; assume it is moving in a circle). Therefore the energy needed to separate the proton and electron is -(-27.2 eV eV) = 13.6 eV. (Although it is classical, this calculation actually gives the correct ionization energy).

23-9 Cathode Ray Tube: TV and Computer Monitors, Oscilloscope
A cathode ray tube contains a wire cathode that, when heated, emits electrons. A voltage source causes the electrons to travel to the anode. Figure If the cathode inside the evacuated glass tube is heated to glowing, negatively charged “cathode rays” (electrons) are “boiled off” and flow across to the anode (+) to which they are attracted.

The electrons can be steered using electric or magnetic fields. Figure A cathode ray tube. Magnetic deflection coils are often used in place of the electric deflection plates shown here. The relative positions of the elements have been exaggerated for clarity.

Televisions and computer monitors (except for LCD and plasma models) have a large cathode ray tube as their display. Variations in the field steer the electrons on their way to the screen. Figure Electron beam sweeps across a television screen in a succession of horizontal lines. Each horizontal sweep is made by varying the voltage on the horizontal deflection plates. Then the electron beam is moved down a short distance by a change in voltage on the vertical deflection plates, and the process is repeated.

An oscilloscope displays an electrical signal on a screen, using it to deflect the beam vertically while it sweeps horizontally. Figure An electrocardiogram (ECG) trace displayed on a CRT.

Review Questions

ConcepTest 23.1a Electric Potential Energy I
A proton and an electron are in a constant electric field created by oppositely charged plates. You release the proton from the positive side and the electron from the negative side. Which feels the larger electric force? A) proton B) electron C) both feel the same force D) neither – there is no force E) they feel the same magnitude force but opposite direction Answer: 5 Electron electron - + Proton proton

ConcepTest 23.1a Electric Potential Energy I
A proton and an electron are in a constant electric field created by oppositely charged plates. You release the proton from the positive side and the electron from the negative side. Which feels the larger electric force? A) proton B) electron C) both feel the same force D) neither – there is no force E) they feel the same magnitude force but opposite direction Since F = qE and the proton and electron have the same charge in magnitude, they both experience the same force. However, the forces point in opposite directions because the proton and electron are oppositely charged. Electron electron - + Proton proton

ConcepTest 23.1b Electric Potential Energy II
A proton and an electron are in a constant electric field created by oppositely charged plates. You release the proton from the positive side and the electron from the negative side. Which has the larger acceleration? A) proton B) electron C) both feel the same acceleration D) neither – there is no acceleration E) they feel the same magnitude acceleration but opposite direction Answer: 2 Electron electron - + Proton proton

ConcepTest 23.1b Electric Potential Energy II
A proton and an electron are in a constant electric field created by oppositely charged plates. You release the proton from the positive side and the electron from the negative side. Which has the larger acceleration? A) proton B) electron C) both feel the same acceleration D) neither – there is no acceleration E) they feel the same magnitude acceleration but opposite direction Since F = ma and the electron is much less massive than the proton, the electron experiences the larger acceleration. Electron electron - + Proton proton

ConcepTest 23.1c Electric Potential Energy III
A proton and an electron are in a constant electric field created by oppositely charged plates. You release the proton from the positive side and the electron from the negative side. When it strikes the opposite plate, which one has more KE? A) proton B) electron C) both acquire the same KE D) neither – there is no change of KE E) they both acquire the same KE but with opposite signs Answer: 3 Electron electron - + Proton proton

ConcepTest 23.1c Electric Potential Energy III
A proton and an electron are in a constant electric field created by oppositely charged plates. You release the proton from the positive side and the electron from the negative side. When it strikes the opposite plate, which one has more KE? A) proton B) electron C) both acquire the same KE D) neither – there is no change of KE E) they both acquire the same KE but with opposite signs Since PE = qV and the proton and electron have the same charge in magnitude, they both have the same electric potential energy initially. Because energy is conserved, they both must have the same kinetic energy after they reach the opposite plate. Electron electron - + Proton proton

ConcepTest 23.2 Work and Potential Energy
Which group of charges took more work to bring together from a very large initial distance apart? 2 +1 d +1 +2 d 1 3 Both took the same amount of work. Answer: 2

ConcepTest 23.2 Work and Potential Energy
Which group of charges took more work to bring together from a very large initial distance apart? 2 +1 d +1 +2 d 1 3 Both took the same amount of work. For case 1: only 1 pair The work needed to assemble a collection of charges is the same as the total PE of those charges: For case 2: there are 3 pairs added over all pairs

ConcepTest 23.3a Electric Potential I
A) V > 0 B) V = 0 C) V < 0 What is the electric potential at point A? A B Answer: 1

ConcepTest 23.3a Electric Potential I
A) V > 0 B) V = 0 C) V < 0 What is the electric potential at point A? A B Since Q2 (which is positive) is closer to point A than Q1 (which is negative) and since the total potential is equal to V1 + V2, the total potential is positive.

ConcepTest 23.3b Electric Potential II
A) V > 0 B) V = 0 C) V < 0 What is the electric potential at point B? A B Answer: 2

ConcepTest 23.3b Electric Potential II
A) V > 0 B) V = 0 C) V < 0 What is the electric potential at point B? A B Since Q2 and Q1 are equidistant from point B, and since they have equal and opposite charges, the total potential is zero. Follow-up: What is the potential at the origin of the x y axes?

ConcepTest 23.4 Hollywood Square
A) E = V = 0 B) E = V  0 C) E  V  0 D) E  V = 0 E) E = V regardless of the value Four point charges are arranged at the corners of a square. Find the electric field E and the potential V at the center of the square. -Q +Q Answer: 4

ConcepTest 23.4 Hollywood Square
A) E = V = 0 B) E = V  0 C) E  V  0 D) E  V = 0 E) E = V regardless of the value Four point charges are arranged at the corners of a square. Find the electric field E and the potential V at the center of the square. The potential is zero: the scalar contributions from the two positive charges cancel the two minus charges. However, the contributions from the electric field add up as vectors, and they do not cancel (so it is non-zero). -Q +Q Follow-up: What is the direction of the electric field at the center?

ConcepTest 23.5a Equipotential Surfaces I
1 3 2 4 +Q –Q At which point does V = 0? E) all of them Answer: 5

ConcepTest 23.5a Equipotential Surfaces I
1 3 2 4 +Q –Q At which point does V = 0? E) all of them All of the points are equidistant from both charges. Since the charges are equal and opposite, their contributions to the potential cancel out everywhere along the mid-plane between the charges. Follow-up: What is the direction of the electric field at all 4 points?

ConcepTest 23.5b Equipotential Surfaces II
Which of these configurations gives V = 0 at all points on the x axis? A) x +2mC -2mC +1mC -1mC B) C) D) all of the above E) none of the above Answer: 1

ConcepTest 23.5b Equipotential Surfaces II
Which of these configurations gives V = 0 at all points on the x axis? A) x +2mC -2mC +1mC -1mC B) C) D) all of the above E) none of the above Only in case (A), where opposite charges lie directly across the x axis from each other, do the potentials from the two charges above the x axis cancel the ones below the x axis.

ConcepTest 23.5c Equipotential Surfaces III
Which of these configurations gives V = 0 at all points on the y axis? A) x +2mC -2mC +1mC -1mC B) C) D) all of the above E) none of the above Answer: 3

ConcepTest 23.5c Equipotential Surfaces III
Which of these configurations gives V = 0 at all points on the y axis? A) x +2mC -2mC +1mC -1mC B) C) D) all of the above E) none of the above Only in case (C), where opposite charges lie directly across the y axis from each other, do the potentials from the two charges above the y axis cancel the ones below the y axis. Follow-up: Where is V = 0 for configuration #2?

ConcepTest 23.6 Equipotential of Point Charge
A) A and C B) B and E C) B and D D) C and E E) no pair Which two points have the same potential? A C B D E Q Answer: 4

ConcepTest 23.6 Equipotential of Point Charge
A) A and C B) B and E C) B and D D) C and E E) no pair Which two points have the same potential? Since the potential of a point charge is: only points that are at the same distance from charge Q are at the same potential. This is true for points C and E. They lie on an equipotential surface. A C B D E Q Follow-up: Which point has the smallest potential?

ConcepTest 23.7a Work and Electric Potential I
A) P  1 B) P  2 C) P  3 D) P  4 E) all require the same amount of work Which requires the most work, to move a positive charge from P to points 1, 2, 3 or 4 ? All points are the same distance from P. P 1 2 3 4 Answer: 1

ConcepTest 23.7a Work and Electric Potential I
A) P  1 B) P  2 C) P  3 D) P  4 E) all require the same amount of work Which requires the most work, to move a positive charge from P to points 1, 2, 3 or 4 ? All points are the same distance from P. P 1 2 3 4 For path #1, you have to push the positive charge against the E field, which is hard to do. By contrast, path #4 is the easiest, since the field does all the work.

ConcepTest 23.7b Work and Electric Potential II
A) P  1 B) P  2 C) P  3 D) P  4 E) all require the same amount of work Which requires zero work, to move a positive charge from P to points 1, 2, 3 or 4 ? All points are the same distance from P. P 1 2 3 4 Answer: 3

ConcepTest 23.7b Work and Electric Potential II
A) P  1 B) P  2 C) P  3 D) P  4 E) all require the same amount of work Which requires zero work, to move a positive charge from P to points 1, 2, 3 or 4 ? All points are the same distance from P. For path #3, you are moving in a direction perpendicular to the field lines. This means you are moving along an equipotential, which requires no work (by definition). P 1 2 3 4 Follow-up: Which path requires the least work?

Ch. 23 Assignments Group problems #’s 4, 10, 12, 30, 34, 36, 42
Homework problems #’s 1, 9, 11, 13, 15, 27, 29, 33, 49, 51, 57 Finish reading Ch. 23 and begin looking over Ch. 24.

Chapter 23 Electric Potential

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